UVa 1611 Crane (构造+贪心)

题意：给定一个序列，让你经过不超过9的6次方次操作，变成一个有序的，操作只有在一个连续区间，交换前一半和后一半。

析：这是一个构造题，我们可以对第 i 个位置找 i 在哪，假设 i  在pos 位置，那么如果 (pos-i)*2+i-1 <= n，那么可以操作一次换过来，

如果不行再换一种，如果他们之间元素是偶数，那么交换 i - pos，如果是奇数，交换 i - pos+1,然后再经过一次就可以换到指定位置。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int pos[maxn], a[maxn];

void Swap(int l, int r){
    int mid = (r-l+1)/2;
    for(int i = 0; i < mid; ++i){
        swap(pos[a[l]], pos[a[l+mid]]);
        swap(a[l], a[l+mid]);
        ++l;
    }
}
vector<P> ans;

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i){
            scanf("%d", a+i);
            pos[a[i]] = i;
        }

        ans.clear();
        for(int i = 1; i <= n; ++i){
            int id = pos[i];
            if(i == a[i])  continue;
            if((id-i)*2+i-1 <= n){
                Swap(i, (id-i)*2+i-1);
                ans.push_back(P(i, (id-i)*2+i-1));
            }
            else if((id-i) & 1){
                Swap(i, id);
                ans.push_back(P(i, id));
            }
            else if(id == n){
                Swap(n-1, n);
                ans.push_back(P(n-1, n));
            }
            else{
                Swap(i, id+1);
                ans.push_back(P(i, id+1));
            }
            --i;
        }
        printf("%d\n", ans.size());
        for(int i = 0; i < ans.size(); ++i)
            printf("%d %d\n", ans[i].first, ans[i].second);
    }
    return 0;
}